Question: $ F = \left[\begin{array}{rr}2 & 3 \\ 0 & 5 \\ -1 & -1\end{array}\right]$ $ D = \left[\begin{array}{rr}-1 & 0 \\ 4 & 2\end{array}\right]$ What is $ F D$ ?
Solution: Because $ F$ has dimensions $(3\times2)$ and $ D$ has dimensions $(2\times2)$ , the answer matrix will have dimensions $(3\times2)$ $ F D = \left[\begin{array}{rr}{2} & {3} \\ {0} & {5} \\ \color{gray}{-1} & \color{gray}{-1}\end{array}\right] \left[\begin{array}{rr}{-1} & \color{#DF0030}{0} \\ {4} & \color{#DF0030}{2}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ F$ , with the corresponding elements in column $j$ of the second matrix, $ D$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ F$ with the first element in ${\text{column }1}$ of $ D$ , then multiply the second element in ${\text{row }1}$ of $ F$ with the second element in ${\text{column }1}$ of $ D$ , and so on. Add the products together. $ \left[\begin{array}{rr}{2}\cdot{-1}+{3}\cdot{4} & ? \\ ? & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ F$ with the corresponding elements in ${\text{column }1}$ of $ D$ and add the products together. $ \left[\begin{array}{rr}{2}\cdot{-1}+{3}\cdot{4} & ? \\ {0}\cdot{-1}+{5}\cdot{4} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ F$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ D$ and add the products together. $ \left[\begin{array}{rr}{2}\cdot{-1}+{3}\cdot{4} & {2}\cdot\color{#DF0030}{0}+{3}\cdot\color{#DF0030}{2} \\ {0}\cdot{-1}+{5}\cdot{4} & ? \\ ? & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{2}\cdot{-1}+{3}\cdot{4} & {2}\cdot\color{#DF0030}{0}+{3}\cdot\color{#DF0030}{2} \\ {0}\cdot{-1}+{5}\cdot{4} & {0}\cdot\color{#DF0030}{0}+{5}\cdot\color{#DF0030}{2} \\ \color{gray}{-1}\cdot{-1}+\color{gray}{-1}\cdot{4} & \color{gray}{-1}\cdot\color{#DF0030}{0}+\color{gray}{-1}\cdot\color{#DF0030}{2}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}10 & 6 \\ 20 & 10 \\ -3 & -2\end{array}\right] $